Given $\displaystyle sinx = \frac{e^{ix} - e^{-ix}}{2i} $ and $\displaystyle cosx = \frac{e^{ix}+ e^{-ix}}{2} $

where $\displaystyle i = \sqrt{-1} $

show that $\displaystyle sin^{4}x = \frac{3}{8} -\frac{1}{2}cos2x + \frac{1}{8} cos4x $

I have a solution for this, provided by my teacher, but i cannot seem to understand it.

the solution is

$\displaystyle sin^{4}x =( \frac{e^{ix}-e^{-ix}}{2})^{4} $ what happened to the i, in the denominator?

this expression is than expanded out, and looks like this; which i dont know how to get to

$\displaystyle \frac{1}{16} ( e^{2ix} + 4e^{2ix} +6 - 4e^{-2ix} + e^{-4ix} $

any help appreciated, thank you.